Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class Main {
	static int n, m;
	static int[][] input;

	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st;
		StringBuilder sb = new StringBuilder();
		st = new StringTokenizer(br.readLine());
		n = Integer.parseInt(st.nextToken());
		m = Integer.parseInt(st.nextToken());
		input = new int[n][m];
		for (int i = 0; i < n; i++) {
			st = new StringTokenizer(br.readLine());
			for (int j = 0; j < m; j++) {
				input[i][j] = Integer.parseInt(st.nextToken());
			}
		}
		int ans = 0;
		while (true) {
			if (checkBad()) {
				System.out.println(-1);
				return;
			}
			if (n > m) {
				if (checkColumns()) {
					for (int i = 0; i < m; i++) {
						int t = input[0][i];
						while (t-- > 0) {
							sb.append("col ").append(i + 1).append('\n');
							ans++;
						}
					}
					break;
				} else if (checkRows()) {
					int max = 0;
					for (int i = 1; i < n; i++) {
						if (input[i][0] > input[max][0])
							max = i;

					}
					sb.append("row ").append(max + 1).append('\n');
					for (int i = 0; i < m; i++) {
						input[max][i] --;
					}
					ans++;
				} else {
					int xmax = 0;
					int ymax = 0;
					for (int i = 0; i < n; i++) {
						for (int j = 0; j < m; j++) {
							if (input[i][j] > input[xmax][ymax]) {
								xmax = i;
								ymax = j;
							}
						}
					}
					sb.append("row ").append(xmax + 1).append('\n');
					sb.append("col ").append(ymax + 1).append('\n');
					ans += 2;
					for (int i = 0; i < n; i++) {
						input[i][ymax]--;
					}
					for (int i = 0; i < m; i++) {
						input[xmax][i]--;
					}
				}
			} else {
				if (checkRows()) {

					for (int i = 0; i < n; i++) {
						int t = input[i][0];
						while (t-- > 0) {
							sb.append("row ").append(i + 1).append('\n');
							ans++;
						}
					}
					break;
				} else if (checkColumns()) {
					int max = 0;
					for (int i = 1; i < m; i++) {
						if (input[0][i] > input[0][max])
							max = i;

					}
					sb.append("col ").append(max + 1).append('\n');
					ans++;
					for (int i = 0; i < n; i++) {
						input[i][max] --;
					}
				} else {
					int xmax = 0;
					int ymax = 0;
					for (int i = 0; i < n; i++) {
						for (int j = 0; j < m; j++) {
							if (input[i][j] > input[xmax][ymax]) {
								xmax = i;
								ymax = j;
							}
						}
					}
					sb.append("row ").append(xmax + 1).append('\n');
					sb.append("col ").append(ymax + 1).append('\n');
					ans += 2;
					for (int i = 0; i < n; i++) {
						input[i][ymax]--;
					}
					for (int i = 0; i < m; i++) {
						input[xmax][i]--;
					}
				}
			}
		}
		System.out.println(ans);
		System.out.print(sb);

	}

	private static boolean checkRows() {
		for (int i = 0; i < n; i++) {
			int v = input[i][0];
			for (int j = 1; j < m; j++) {
				if (input[i][j] != v)
					return false;
			}
		}
		return true;
	}

	private static boolean checkColumns() {
		for (int i = 0; i < m; i++) {
			int v = input[0][i];
			for (int j = 1; j < n; j++) {
				if (input[j][i] != v)
					return false;
			}
		}
		return true;
	}

	private static boolean checkBad() {
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < m; j++) {
				if (input[i][j] < 0)
					return true;
			}
		}
		return false;
	}
}